How do you find the equation of the tangent line to the curve #f(x) = (x-1)^3# at the point where x = 2?

Question

Scarlett Allison · Accepted Answer

y = 3x - 5

To find the equation in the form y = mx + c , where m represents the gradient and c, the y-intercept. The value of f'(2) will give m and f(2) will assist in finding c. differentiate using the #color(blue)" chain rule "# #d/dx f(g(x)) = f'(g(x)) . g'(x)# hence f'(x) #=3(x -1)^2 d/dx(x - 1 ) = 3(x - 1 )^2# f'(2) = #3(1)^2 = 3 " gradient of tangent "# and f(2) =# (1)^3 = 1 rArr (2,1) " is tangent point " # partial equation of tangent is y = 3x + c , and using (2,1 ) gives : 1 =3(2) + c → c = - 5 thus equation of tangent is : y = 3x - 5

William Abdalla · Answer

undefined To find the equation of the tangent line to the curve f(x) = (x-1)^3 at the point where x = 2, we need to find the slope of the tangent line and the coordinates of the point of tangency.

First, we find the derivative of f(x) with respect to x, which gives us f'(x) = 3(x-1)^2.

Next, we substitute x = 2 into f'(x) to find the slope of the tangent line at x = 2. This gives us f'(2) = 3(2-1)^2 = 3.

Now, we have the slope of the tangent line, which is 3, and the point of tangency, which is (2, f(2)). To find f(2), we substitute x = 2 into f(x), giving us f(2) = (2-1)^3 = 1.

Therefore, the point of tangency is (2, 1), and the slope of the tangent line is 3.

Using the point-slope form of a linear equation, y - y1 = m(x - x1), where (x1, y1) is the point of tangency and m is the slope, we can substitute the values to find the equation of the tangent line.

Substituting (2, 1) for (x1, y1) and 3 for m, we have y - 1 = 3(x - 2).

Simplifying this equation gives us the equation of the tangent line to the curve f(x) = (x-1)^3 at the point where x = 2: y = 3x - 5.

Adam Adkins · Answer

undefined To find the equation of the tangent line to the curve $ f(x) = (x-1)^3 $ at the point where $ x = 2 $, follow these steps:

1. Find the derivative of $ f(x) $ using the power rule: $ f'(x) = 3(x-1)^2 $.
2. Evaluate $ f'(2) $ to find the slope of the tangent line at $ x = 2 $: $ f'(2) = 3(2-1)^2 = 3 $.
3. Use the point-slope form of the equation of a line: $ y - y_1 = m(x - x_1) $, where $ (x_1, y_1) $ is the point of tangency and $ m $ is the slope.
4. Substitute $ x_1 = 2 $, $ y_1 = f(2) $, and $ m = f'(2) $ into the point-slope form equation.
5. Simplify the equation to find the tangent line's equation.

Therefore, the equation of the tangent line to the curve $ f(x) = (x-1)^3 $ at the point where $ x = 2 $ is $ y - 1 = 3(x - 2) $.